How would I solve: $\log_{16} 32 = x$?

Question

How would I solve: $\log_{16} 32 = x$?

What I know:

• 16 is the base
  
• 32 is the exponent

$$ 32 = 16^x $$

I'm stuck at this point$\ldots$

Answer 1

$$32=16^x$$

Hint:

$$2^5=2^{4x}$$

Spoiler:

$$4x=5$$ $$\boxed{x=\frac 5 4}$$

Answer 2

Hint. Observe that $32 = 16 \cdot 2$, and $16 = 2^4$, so $2 = 16^z$ where $z = $ what?

Answer 3

First, i want to point out that $x=\log_{16}32$ IS a solution, just not a very useful one since you need to convert to base $e$ or base $10$ to use a calculator usually.

Now that you have $16^x=32$, you can use a log of known base (most commonly, natural log) and plug that in to a calculator to find a value.

$\ln(16^x)=\ln32$

$x\ln16=\ln32$

$x=\frac{\ln32}{\ln16}$

Answer 4

$32 = 16^x$ is the same as $32 = (2^4)^x = 2^{4x}$ and you know that $32 = 2^5$ and so $32 = 2^\color{red}{5} = 2^{\color{red}{4x}}$and so $\color{red}{4x = 5}$ and hence $$x = \frac{5}{4}$$